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Topic: zeta functions for P-smooth numbers (?)
Replies: 9   Last Post: Oct 2, 2011 5:17 AM

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 David Bernier Posts: 3,892 Registered: 12/13/04
Re: zeta functions for P-smooth numbers (?)
Posted: Mar 14, 2011 2:56 PM

David Bernier wrote:
> Suppose P is some infinite subset of the primes {2, 3, 5, 7, ... }.
> A positive integer n is called P-smooth if and only if all its prime
> factors belong to P.
>
> For s in C with Re(s)> 1, one defines a "zeta_P" function as follows:
>
> zeta_P (s) := sum_{n >= 1 and n a P-smooth number} n^(-s) .
>
> I'm wondering if one has a valid Euler-product such as:
>
>
> sum_{n >= 1 and n a P-smooth number} n^(-s) = prod_{p in P} 1/(1-p^(-s)) ?
>
> Would there be a zeta_P functional equation?
>
> Would there be a functional equation?
>
> For a specific example of a P with a very sparse complement relative to
> the primes, we might consider the very sparse infinite set of primes
>
> T defined by:
>
> 1- Let c_1 be the least prime greater than Graham's number G with G as
> referenced in the Wikipedia article on Graham's number:
> < http://en.wikipedia.org/wiki/Graham's_number >
> refers to an article on Ramsey theory by Graham & Rothschild (1971)
> and a number " G = f^64 (4)" which was described in Martin Gardner's
> Mathematical Games column for November 1977.
> Wikipedia:
> << This weaker upper bound [i.e. G as above] , attributed to some
> unpublished work of Graham, was eventually published (and dubbed
> Graham's number) by Martin Gardner, in [Scientific American,
> "Mathematical Games", November 1977]. >>
>
>
> 2- If c_k has been defined, k >=1, let
> c_{k+1} := the least prime greater than 2^(c_k) .
>
> Then the sparse infinite set of primes T is defined by
> T = { natural numbers n such that n=c_k for some k in N^* . }.
>
> So T is a very sparse infinite set of primes. Now we can let
> P = Primes \ T , '\' denoting set-theoretical difference, and
> Primes being the set of positive integers greater than 1
> with exactly two (positive) divisors.
>
> Discussion numbers that aren't P-smooth include c_1, (c_1)^2, (c_1)*(c_2) ,
> and so on.
>
> I recall below the definition of zeta_P :
>
> ==> zeta_P (s) := sum_{n >= 1 and n a P-smooth number} n^(-s) . [ Re(s)

> > 1]
>
> Following Harold Edwards book "Riemann's Zeta Function",
> how much of the plan of (roughly) Chapters 1 through 7 could
> be carried out?
>
> ---
>
> Motivation: this question of a form of zeta function for P-smooth numbers
> that I call "zeta-P" interests me because the infinite set
> P can be made to-order; for example, one could have a subset
> of the Primes P such that the "primes in P"-counting function
> say pi_P (.) would satisfy Schoenfeld's explicit
> version of von Koch's Theorem as long as x < G,
> G being Graham's number:
> http://en.wikipedia.org/wiki/Riemann_hypothesis#Distribution_of_prime_numbers
>
>
> for x > G, one could have for arbitrarily large x:
> | pi_P (x) - Li(x)| > x^(51/100)
> while still respecting | pi_P (x) - Li(x)| = O(x^(51/100+epsilon)),
> for any given epsilon > 0.
> NB.: pi_P (x) is a count of the primes p belonging to P and less than or
> equal to x .

[...]

One of the simplest cases, in the same spirit, is to relax the
condition on the cardinality of Primes \ P, by allowing
Primes \ P to have a finite cardinality. One might as
well first try to remove just one prime; if one is
to remove just one prime, one can try removing the smallest, 2.

Then P = Primes \ {2}. The P-smooth integers are the odd numbers
{1, 3, 5, 7, ... }.

zeta_P (s) := 1 + 1/3^s + 1/5^s + 1/7^s + 1/9^s + 1/11^s + ... , Re(s) > 1.

(a) Does zeta_P (s) have an analytic continuation?
(b) What if one tries to perform Euler-Maclaurin summation
on zeta_P(1/2 + it) "=" 1 + 1/[3^(1/2 + it)] + 1/[5^(1/2 + it)]
+ 1/[7^(1/2 + it)]+ 1/[9^(1/2 + it)] + ...
for t > 0 ? analytic? convergent?
where are the zeros? Does one have an analog for
von Mangoldt's explicit formula for the summatory Lambda function,
psi(x) in Edwards, section 3.2, page 54 ? etc.

David Bernier

Date Subject Author
3/14/11 David Bernier
3/18/11 David Bernier
3/18/11 David Bernier
3/20/11 David Bernier
9/25/11 David Bernier
9/30/11 David Bernier
10/1/11 David Bernier
10/2/11 David Bernier
10/2/11 David Bernier